The graph for this equation should have a hole at origin, since we can't just divide 1 by 0. y/x is the golden ratio number or its conjugate, both of them irrational, hence no rational points.
We can solve for y by multiplying everything by the LCD, which is xy(x+y). That equation is y(x+y)-x(x+y)=xy. Using distributive property and moving one side to the other changes this as xy+y^2-x^2-xy-xy=0. Rearranging in terms of y yields the equation as y^2-xy-x^2=0. You can solve for y by using the quadratic formula, so y=(-(-x)±√((-x)^2-4(1)(-x^2)))/(2(1)). Simplifying this expression is (x±√(x^2+4x^2))/2, or (1±√(5))x/2. That graph has two lines that looks like it's going to intersect at x=o, but it is not there at that point and should have a hole to show a point of discontinuity.
It is also interesting to note that this equation is related to a conic section (ellipse, parabola, hyperbola) which has been rotated about the origin through an angle t. The presence of the xy term indicates the rotation. The General Form is : ax^2 + 2hxy + by^2 = k. Without going into the details, you can determine the angle, t, that the conic has been rotated through by solving: tan(2t) = 2h/(b-a) and also determine what type of conic section you have by evaluating the expression : ab - h^2. ab - h^2 > 0 ===> ellipse ab - h^2 = 0 ===> parabola ab - h^2 hyperbola. So: x^2 + xy - y^2 = k has a=1, h=1/2,b=-1. Note the k and not 0. I'll get to that. These values of a, b, h give us ab - h^2 = (1)(-1) - (1/2)^2 = -5/4 indicating a hyperbola. If you replace the k with a 0 you get what is called a degenerate hyperbola. A hyperbola has 2 straight line asymptotes. As k ---> 0 the hyperbola degenerates into these 2 straight lines. So x^2 + xy - y^2 = 0 is a degenerate hyperbola. i.e the 2 straight line asymptotes you have given as your solution. Cheers.
@@SyberMath Thanks. I taught this for many years in High School in Ontario. Here's what my students could do as a standard "put it all together" problem. Try it as a challenge. Place the following conic in "standard position" by applying a rotation followed by a translation. Use the transformations you have found to find the foci of this equation. 5x^2 + 8xy + 5y^2 + 13[sqrt(2)]x + 5[sqrt(2)]y = -16. So you can check: Foci are: ( 2 - 5/sqrt(2) , -2 + 3/sqrt(2) ) and ( -2 - 5/sqrt(2) , 3 + 3/sqrt(2) ) The two 5's might mislead you into thinking this is a circle. It isn't.
But what about the point of origin (0,0)? This point satisfies both the equations but can they be a solution to the original equation? I mean will it satisfy 1/x - 1/y = 1/(x+y)?
A fun and *golden* rational equation! 😄
Thank you!
The graph for this equation should have a hole at origin, since we can't just divide 1 by 0.
y/x is the golden ratio number or its conjugate, both of them irrational, hence no rational points.
We can solve for y by multiplying everything by the LCD, which is xy(x+y). That equation is y(x+y)-x(x+y)=xy. Using distributive property and moving one side to the other changes this as xy+y^2-x^2-xy-xy=0. Rearranging in terms of y yields the equation as y^2-xy-x^2=0. You can solve for y by using the quadratic formula, so y=(-(-x)±√((-x)^2-4(1)(-x^2)))/(2(1)). Simplifying this expression is (x±√(x^2+4x^2))/2, or (1±√(5))x/2. That graph has two lines that looks like it's going to intersect at x=o, but it is not there at that point and should have a hole to show a point of discontinuity.
It is also interesting to note that this equation is related to a conic section (ellipse, parabola, hyperbola) which has been rotated about the origin through an angle t. The presence of the xy term indicates the rotation. The General Form is : ax^2 + 2hxy + by^2 = k. Without going into the details, you can determine the angle, t, that the conic has been rotated through by solving: tan(2t) = 2h/(b-a)
and also determine what type of conic section you have by evaluating the expression : ab - h^2.
ab - h^2 > 0 ===> ellipse
ab - h^2 = 0 ===> parabola
ab - h^2 hyperbola.
So: x^2 + xy - y^2 = k has a=1, h=1/2,b=-1. Note the k and not 0. I'll get to that. These values of a, b, h give us ab - h^2 = (1)(-1) - (1/2)^2 = -5/4 indicating a hyperbola. If you replace the k with a 0 you get what is called a degenerate hyperbola. A hyperbola has 2 straight line asymptotes. As k ---> 0 the hyperbola degenerates into these 2 straight lines.
So x^2 + xy - y^2 = 0 is a degenerate hyperbola. i.e the 2 straight line asymptotes you have given as your solution.
Cheers.
Wow, nice work! 🤩
@@SyberMath Thanks. I taught this for many years in High School in Ontario. Here's what my students could do as a standard "put it all together" problem. Try it as a challenge.
Place the following conic in "standard position" by applying a rotation followed by a translation. Use the transformations you have found to find the foci of this equation.
5x^2 + 8xy + 5y^2 + 13[sqrt(2)]x + 5[sqrt(2)]y = -16. So you can check: Foci are: ( 2 - 5/sqrt(2) , -2 + 3/sqrt(2) ) and ( -2 - 5/sqrt(2) , 3 + 3/sqrt(2) )
The two 5's might mislead you into thinking this is a circle. It isn't.
@@ianfowler9340 is it related to conic section?
I love these!
Glad to hear that! 🤩
Good luck 🍀
Thank you 😁
Very nice! I used the second method!!!!❤❤❤
Great!!
Love your videos! Keep up the great work! 👍🏻
Thank you! Will do!
Should the graph have a circle at the origin since x ≠ 0 and y ≠ 0?
(:
yes but desmos does not show open dots as far as I know
That was golden
also notice how the lines are perpendicular to each other because of the golden ratio
A Golden Intersection! 🤩
Nice golden video!
Thanks, Juan!
Cool - I used the second method, except I didn't substitute.
Great job!
Multiplying both sides with x+y we get y/x-x/y=1 or (y/x)^2-(y/x)-1=0 which can be solved with quadratic formula.
Here is a slightly shorter solution:
1/x+1/y=x+y
y+x=xy(x+y)
y+x-xy(x+y)=0
(x+y)(1-xy)=0
x+y=0 or 1-xy=0
y=-x or y=1/x where x≠0 and y≠0
But what about the point of origin (0,0)? This point satisfies both the equations but can they be a solution to the original equation? I mean will it satisfy 1/x - 1/y = 1/(x+y)?
The solution is number Aureo
Fun! Great! =)
Thanks! 😃
Nice results！😏😏😏😏😏😏